# Homework 2: Higher Order Functions

*Due by 11:59pm on Thursday, July 7*

## Instructions

Download hw02.zip. Inside the archive, you will find
a file called hw02.py, along with a copy of the `ok`

autograder.

**Submission:** When you are done, submit with ```
python3 ok
--submit
```

. You may submit more than once before the deadline; only the
final submission will be scored. Check that you have successfully submitted
your code on okpy.org. See Lab 0 for more instructions on
submitting assignments.

**Using Ok:** If you have any questions about using Ok, please
refer to this guide.

**Readings:** You might find the following references
useful:

**Grading:** Homework is graded based on
correctness. Each incorrect problem will decrease the total score by one point. There is a homework recovery policy as stated in the syllabus.
**This homework is out of 2 points.**

# Required questions

Several doctests refer to these functions:

```
from operator import add, mul
square = lambda x: x * x
identity = lambda x: x
triple = lambda x: 3 * x
increment = lambda x: x + 1
```

## Getting Started Videos

These videos may provide some helpful direction for tackling the coding problems on this assignment.

To see these videos, you should be logged into your berkeley.edu email.

### Q1: Product

Write a function called `product`

that returns `term(1) * ... * term(n)`

.

```
def product(n, term):
"""Return the product of the first n terms in a sequence.
n: a positive integer
term: a function that takes one argument to produce the term
>>> product(3, identity) # 1 * 2 * 3
6
>>> product(5, identity) # 1 * 2 * 3 * 4 * 5
120
>>> product(3, square) # 1^2 * 2^2 * 3^2
36
>>> product(5, square) # 1^2 * 2^2 * 3^2 * 4^2 * 5^2
14400
>>> product(3, increment) # (1+1) * (2+1) * (3+1)
24
>>> product(3, triple) # 1*3 * 2*3 * 3*3
162
"""
"*** YOUR CODE HERE ***"
```

Use Ok to test your code:

`python3 ok -q product`

### Q2: Accumulate

Let's take a look at how `product`

is an instance of a more
general function called `accumulate`

, which we would like to implement:

```
def accumulate(merger, start, n, term):
"""Return the result of merging the first n terms in a sequence and start.
The terms to be merged are term(1), term(2), ..., term(n). merger is a
two-argument commutative function.
>>> accumulate(add, 0, 5, identity) # 0 + 1 + 2 + 3 + 4 + 5
15
>>> accumulate(add, 11, 5, identity) # 11 + 1 + 2 + 3 + 4 + 5
26
>>> accumulate(add, 11, 0, identity) # 11
11
>>> accumulate(add, 11, 3, square) # 11 + 1^2 + 2^2 + 3^2
25
>>> accumulate(mul, 2, 3, square) # 2 * 1^2 * 2^2 * 3^2
72
>>> # 2 + (1^2 + 1) + (2^2 + 1) + (3^2 + 1)
>>> accumulate(lambda x, y: x + y + 1, 2, 3, square)
19
>>> # ((2 * 1^2 * 2) * 2^2 * 2) * 3^2 * 2
>>> accumulate(lambda x, y: 2 * x * y, 2, 3, square)
576
>>> accumulate(lambda x, y: (x + y) % 17, 19, 20, square)
16
"""
"*** YOUR CODE HERE ***"
```

`accumulate`

has the following parameters:

`term`

and`n`

: the same parameters as in`product`

`merger`

: a two-argument function that specifies how the current term is merged with the previously accumulated terms.`start`

: value at which to start the accumulation.

For example, the result of `accumulate(add, 11, 3, square)`

is

`11 + square(1) + square(2) + square(3) = 25`

Note:You may assume that`merger`

is commutative. That is,`merger(a, b) == merger(b, a)`

for all`a`

,`b`

, and`c`

. However, you may not assume`merger`

is chosen from a fixed function set and hard-code the solution.

After implementing `accumulate`

, show how `summation`

and `product`

can both be
defined as function calls to `accumulate`

.

**Important:**
You should have a single line of code (which should be a `return`

statement)
in each of your implementations for `summation_using_accumulate`

and
`product_using_accumulate`

, which the syntax check will check for.

```
def summation_using_accumulate(n, term):
"""Returns the sum: term(0) + ... + term(n), using accumulate.
>>> summation_using_accumulate(5, square)
55
>>> summation_using_accumulate(5, triple)
45
>>> # You aren't expected to understand the code of this test.
>>> # Check that the bodies of the functions are just return statements.
>>> # If this errors, make sure you have removed the "***YOUR CODE HERE***".
>>> import inspect, ast
>>> [type(x).__name__ for x in ast.parse(inspect.getsource(summation_using_accumulate)).body[0].body]
['Expr', 'Return']
"""
"*** YOUR CODE HERE ***"
def product_using_accumulate(n, term):
"""Returns the product: term(1) * ... * term(n), using accumulate.
>>> product_using_accumulate(4, square)
576
>>> product_using_accumulate(6, triple)
524880
>>> # You aren't expected to understand the code of this test.
>>> # Check that the bodies of the functions are just return statements.
>>> # If this errors, make sure you have removed the "***YOUR CODE HERE***".
>>> import inspect, ast
>>> [type(x).__name__ for x in ast.parse(inspect.getsource(product_using_accumulate)).body[0].body]
['Expr', 'Return']
"""
"*** YOUR CODE HERE ***"
```

Use Ok to test your code:

```
python3 ok -q accumulate
python3 ok -q summation_using_accumulate
python3 ok -q product_using_accumulate
```

### Q3: Filtered Accumulate

Extend the `accumulate`

function from the previous question to allow for *filtering* the results
produced by its `term`

argument by filling in the implementation for the
`filtered_accumulate`

function:

```
def filtered_accumulate(merger, start, cond, n, term):
"""Return the result of merging the terms in a sequence of N terms
that satisfy the condition cond. merger is a two-argument function.
If v1, v2, ..., vk are the values in term(1), term(2), ..., term(N)
that satisfy cond, then the result is
start merger v1 merger v2 ... merger vk
(treating merger as if it were a binary operator, like +). The
implementation uses accumulate.
>>> filtered_accumulate(add, 0, lambda x: True, 5, identity) # 0 + 1 + 2 + 3 + 4 + 5
15
>>> filtered_accumulate(add, 11, lambda x: False, 5, identity) # 11
11
>>> filtered_accumulate(add, 0, odd, 5, identity) # 0 + 1 + 3 + 5
9
>>> filtered_accumulate(mul, 1, greater_than_5, 5, square) # 1 * 9 * 16 * 25
3600
>>> # Do not use while/for loops or recursion
>>> from construct_check import check
>>> # ban iteration and recursion
>>> check(HW_SOURCE_FILE, 'filtered_accumulate', ['While', 'For', 'Recursion'])
True
"""
def merge_if(x, y):
"*** YOUR CODE HERE ***"
return accumulate(merge_if, start, n, term)
def odd(x):
return x % 2 == 1
def greater_than_5(x):
return x > 5
```

`filtered_accumulate`

has the following parameters:

`merger`

,`start`

,`n`

and`term`

: the same arguments as`accumulate`

.`cond`

: a one-argument function applied to the values of`term(k)`

for each`k`

from 1 to`n`

. Only values for which`cond`

returns a true value are included in the accumulated total. If no values satisfy`cond`

, then`start`

is returned.

For example, the result of ```
filtered_accumulate(add, 0, is_prime, 11,
identity)
```

is

`0 + 2 + 3 + 5 + 7 + 11`

for a valid definition of `is_prime`

.

Implement `filtered_accumulate`

by defining the `merge_if`

function. Exactly
what this function does is something for you to discover. Do not use any loops
or make any recursive calls to `filtered_accumulate`

.

Hint:The order in which you pass the arguments to`merger`

in your solution to`accumulate`

matters here.

Use Ok to test your code:

`python3 ok -q filtered_accumulate`

### Q4: Funception

Write a function (funception) that takes in another function `func_a`

and a number `start`

and returns a function (`func_b`

) that will have
one parameter to take in the stop value. `func_b`

should take the
following into consideration the following in order:

- Takes in the stop value.
- If the value of
`start`

is less than 0, exit the function by returning`None`

. - If the value of
`start`

is greater than stop, apply`func_a`

on`start`

and return the result. - If not, apply
`func_a`

on all the numbers from start (inclusive) up to stop (exclusive) and return the product.

```
def funception(func_a, start):
""" Takes in a function (function A) and a start value.
Returns a function (function B) that will find the product of
function A applied to the range of numbers from
start (inclusive) to stop (exclusive)
>>> def func_a(num):
... return num + 1
>>> func_b1 = funception(func_a, 0)
>>> func_b1(3) # func_a(0) * func_a(1) * func_a(2) = 1 * 2 * 3 = 6
6
>>> func_b2 = funception(func_a, 1)
>>> func_b2(4) # func_a(1) * func_a(2) * func_a(3) = 2 * 3 * 4 = 24
24
>>> func_b3 = funception(func_a, 3)
>>> func_b3(2) # Returns func_a(3) since start > stop
4
>>> func_b4 = funception(func_a, -2)
>>> func_b4(-3) # Returns None since start < 0
>>> func_b5 = funception(func_a, -1)
>>> func_b5(4) # Returns None since start < 0
"""
"*** YOUR CODE HERE ***"
```

Use Ok to test your code:

`python3 ok -q funception`

## Submit

Make sure to submit this assignment by running:

`python3 ok --submit`

## Bonus Questions

Homework assignments will also contain prior exam-level questions for you to take a look at. These questions have no submission component; feel free to attempt them if you'd like a challenge!

Note that exams from Spring 2020, Fall 2020, and Spring 2021 gave students access to an interpreter, so the question format may be different than other years. Regardless, the questions included remain good exam-level problems doable

withoutaccess to an interpreter.

- Fall 2019 MT1 Q3: You Again [Higher Order Functions]
- Spring 2021 MT1 Q4: Domain on the Range [Higher Order Functions]
- Fall 2021 MT1 Q1b: tik [Functions and Expressions]